The unbounded fixed-pivot flow has first-order grid error

For every $z\ge a$, [F6] gives $\int w(u)P_{\mathrm{\Delta }}(z)(du)\le w(z)+{\mathrm{\Delta }}^2\mathrm{/}4\le (5\mathrm{/}4)w(z)$, so the linear projection $P_{\mathrm{\Delta }}$ is bounded on ${\mathcal{M}}_w(E)$. Set $q_t=P_{\mathrm{\Delta }}{\mu }_t$; applying this bounded linear map to the integral equation [F1] gives ${\dot{q}}_t=P_{\mathrm{\Delta }}F({\mu }_t)$. The total-number-one and first-moment identities in [F2] and [F6] give $M_0(q_t)=M_0({\mu }_t)$ and $M_1(q_t)=M_1({\mu }_t)$, so [F5] supplies a $\mathrm{\Delta }$-independent number bound for both $q_t$ and ${\eta }_t$.

Subtract the equations for ${\eta }_t$ and $q_t$ and add and subtract $F_{\mathrm{\Delta }}(q_t)$. The vector-field term is bounded by $C_T{d}_{BL}({\eta }_t,q_t)$ using [F5]; the residual is at most $r:=C_Q\mathrm{\Delta }$ by [F3]. Therefore $d(t)\le C_T{\int }_0^{t}d(s)ds+rt$, with $d(0)=0$. Set $g(t)=rt+C_T{\int }_0^{t}d(s)ds$. Then $d\le g$ and $g^{\mathrm{\prime }}\le r+C_{T}g$. If $C_T>0$, integrating $(e^{-C_{T}t}g(t){)}^{\mathrm{\prime }}\le r{e}^{-C_{T}t}$ gives $g(t)\le r(e^{C_{T}t}-1)\mathrm{/}{C}_T\le rt{e}^{C_{T}t}$; if $C_T=0$, $g(t)=rt$. Since $t\le T$, this is at most $e^{C_{T}t}{C}_{Q}T\mathrm{\Delta }$.

For any test admissible in [F3], its expected pivot value differs from its original value by at most its Lipschitz constant times the expected displacement. [F6] therefore yields $d_{BL}(q_t,{\mu }_t)\le (\mathrm{\Delta }\mathrm{/}2)M_0({\mu }_t)$. The triangle inequality with step 2.1 proves the Statement.